Verified Answer :
Area of a triangle=1/2bh
h=b-5
42=1/2b(b-5)
42=1/2*b^2-5b
84=b^2-5b
b^2-5b-84=0
b^2+7b-12b-84=0
b(b+7)-12(b+7)=0
b=12,b=-7
so width cant be negative so base =12 m
H=12-5=7 m
Answer:
The length of the base is 12 m
Step-by-step explanation:
Given : The height of a triangle is 5 m less than its base. The area of the triangle is 42 m².
We have to find the length of the base.
Given the height of triangle is 5 m less than base
Let base be x then height is ( x -5 ) m
Area of triangle is \(\frac{1}{2} \times base \times height\)
also given area of triangle is 42 m²
substitute, we get,
\(42=\frac{1}{2} \times x \times (x-5)\)
Solving , we get,
\(84=x(x-5)\)
Simplify we get a quadratic equation as,
\(x^2-5x-84=0\)
Using quadratic formula,
For the standard quadratic equation \(ax^2+bx+c=0\) , the solution of roots is given by, \(x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
For a = 1 , b= -5 , c = -84 , we get
\(x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\left(-84\right)}}{2\cdot \:1}\)
Simplify, we get
\(=\frac{5\pm\sqrt{361}}{2\cdot \:1}\)
We know \(\sqrt{361}=19\) , we get
\(x_{1,2}=\frac{5\pm19}{2}\)
\(x_1=\frac{5+19}{2}\) and \(x_2=\frac{5-19}{2}\)
\(x_1=12\) and \(x_2=-7\)
ignoring negative value as side cannot be negative.
thus, x = 12
base is 12 m then height is ( 12 -5 ) m = 4 m
Thus, the length of the base is 12 m.